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How to prove that prove that $S$ is weak$^*$-continuous from $Y^*$ to $X^*$ if an only if $S=T^*$ for some $T\in B(X,Y)$

Thanks for any hints.

To show that $T$ is continuous is straight forward but inverse is hard.

Martin Argerami
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user330305
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    This is ridiculous. A perfectly normal question in functional analysis is closed by 5 users, where four of them have zero reputation in the tag, and the other 330. – Martin Argerami Jun 20 '16 at 04:29
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    Anyway, while we wait for reasonable people to look at this question, here is an answer so I don't lose what I spent 15 minutes typing. If $S=T^$, and $f_j\to f$ weak$^$ in $Y^$, then for any $x\in X$ $$ Sf_j(x)=f_j(Tx)\to f(Tx)=T^f(x)=Sf(x). $$ So $S$ is weak$^*$-continuous.

    (continued)

     – Martin Argerami Jun 20 '16 at 04:34
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    Conversely, assume that $S$ is weak$^$-continuous. We know that what our $T$ should satisfy it is exists: $Sf(x)=f(Tx)$. So let us use this to define $T$. For any $x\in X$, consider the functional on $Y^$ given by $\alpha_x:f\longmapsto Sf(x)$. By hypothesis this is weak$^$-continuous; the weak$^$-continuous functionals are bounded, so $\alpha_x\in Y^{*}$. (continued)* – Martin Argerami Jun 20 '16 at 04:36
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    But, also, a weak$^*$-continuous functional on the dual is in the predual. So there exists $y\in Y$ with $\alpha_x(f)=f(y)$ for all $f\in Y^$. Define $Tx=y$. Note that $y$ is unique because $Y^$ separates points in $Y$. From there we deduce that $T$ is linear. Finally, $$ |Tx|=\sup{|f(Tx)|:\ f\in Y^,\ |f|=1}=\sup{|Sf(x)|:\ f\in Y^,\ |f|=1}\leq|S|,|x|. $$So $T$ is bounded with $|T|\leq|S|$. – Martin Argerami Jun 20 '16 at 04:36
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    @MartinArgerami Agreed. I like this question just fine as is, and expect any attempts to add "context" would just clutter things up and make it less useful for future readers. – Nick Alger Jun 20 '16 at 04:44
  • Thanks Nick. If not enough people appear to reopen it, I'll post it and answer it myself. Meanwhile, another user is scared away from the site, thanks to the closing brigade. – Martin Argerami Jun 20 '16 at 04:47
  • So I already posted the question and answer (I'll delete it if this one gets reopened), so that I keep what I typed. – Martin Argerami Jun 20 '16 at 04:53

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